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The document discusses dynamic programming and its application to solving the longest common subsequence (LCS) problem. It presents the LCS algorithm, which uses dynamic programming to find the length and sequence of the longest subsequence common to two strings X and Y in O(mn) time, where m and n are the lengths of X and Y, respectively. It provides an example running the LCS algorithm on strings X="ABCB" and Y="BDCAB" to determine their longest common subsequence is "BCB".
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Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
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The document reviews concepts related to NP-completeness, including reductions between problems. It provides examples of reducing the directed Hamiltonian cycle problem to the undirected version. It also reduces 3-SAT to the clique problem by transforming a Boolean formula to a graph, then further reduces clique to vertex cover. Hundreds of problems have been shown to be NP-complete through relatively simple reductions like these that leverage previous results.
lecture 29
The document discusses reductions between problems to prove NP-completeness. It first reviews P, NP, reductions, NP-hard and NP-complete problems. It then walks through reducing directed Hamiltonian cycle to undirected Hamiltonian cycle to traveling salesman problem (TSP) to prove that TSP is NP-complete in 3 steps or less.
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This document discusses NP-completeness and algorithms. It begins by reviewing tractable versus intractable problems, and the complexity classes P and NP. P contains problems solvable in polynomial time, while NP includes problems verifiable in polynomial time. The document then discusses NP-complete problems, which are the hardest problems in NP. It notes that reducing one problem to another shows the first problem is no harder than the second. The document concludes by outlining the steps to prove a problem is NP-complete: choosing a known NP-complete problem, reducing it to the target problem in polynomial time, and showing the target is in NP.
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The document discusses NP-completeness and algorithms. It introduces dynamic programming and greedy algorithms. It then discusses the activity selection problem and shows it can be solved greedily by choosing the activity with the earliest finish time at each step. Finally, it defines the classes P and NP, introduces the concept of reductions to show problems are NP-complete, and states that if any NP-complete problem can be solved in polynomial time, then P=NP.
lecture 26
The document discusses various algorithms techniques including greedy algorithms, dynamic programming, and their application to problems like the activity selection problem and knapsack problem. It provides examples of optimal subproblems, overlapping subproblems, and how dynamic programming and greedy algorithms can be used to solve problems exhibiting these properties.
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The document describes the 0-1 knapsack problem and how to solve it using dynamic programming. The 0-1 knapsack problem involves packing items of different weights and values into a knapsack of maximum capacity to maximize the total value without exceeding the weight limit. A dynamic programming algorithm is presented that breaks the problem down into subproblems and uses optimal substructure and overlapping subproblems to arrive at the optimal solution in O(nW) time, improving on the brute force O(2^n) time. An example is shown step-by-step to illustrate the algorithm.
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The document discusses algorithms for solving the single-source shortest path problem, including Dijkstra's algorithm, Bellman-Ford algorithm, and shortest paths in directed acyclic graphs (DAGs). It reviews the key ideas behind relaxation and how algorithms like Bellman-Ford and Dijkstra's use relaxation to iteratively improve the upper bound on shortest path distances. Running times of the algorithms are analyzed, with Dijkstra's algorithm taking O(ElogV) time and Bellman-Ford taking O(VE) time.
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The document describes Prim's algorithm for finding a minimum spanning tree (MST) of a weighted undirected graph. It shows the pseudocode for Prim's algorithm and step-by-step workings on an example graph to build up the MST. It notes that the key operation is decreasing the key value, which is called once for each edge in the MST, and the running time depends on the priority queue implementation, being O(ElgV) for a binary heap and O(VlgV+E) for a Fibonacci heap.
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The document discusses algorithms for finding minimum spanning trees in graphs. It describes two common algorithms: Kruskal's algorithm and Prim's algorithm. Kruskal's algorithm sorts the edges by weight and builds up the spanning tree by adding edges that do not create cycles. Prim's algorithm maintains a set of vertices in the spanning tree and iteratively adds the lowest weight edge connecting a vertex outside the set to one inside. Both run in O((V+E)logV) time using efficient data structures.
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The document discusses two algorithms for matrix multiplication and finding the median of an unsorted list:
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The document discusses binary search trees (BSTs) and their use as a data structure for dynamic sets. It covers BST properties, operations like search, insert, delete and their running times. It also discusses using BSTs to sort an array in O(n lg n) time by inserting elements, similar to quicksort. Maintaining a height of O(lg n) is important for efficient operations.
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This document discusses binary search trees (BSTs) and their use for dynamic sets and sorting. It covers BST operations like search, insert, find minimum/maximum, and delete. It explains that BST sorting runs in O(n log n) time like quicksort. Maintaining a height of O(log n) can lead to efficient implementations of priority queues and other dynamic set applications using BSTs.
lecture 10
The document discusses various sorting algorithms and their time complexities:
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lecture 24
- 11. LCS Example (0) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5 Allocate array c[5,4] ABCB BDCAB
- 12. LCS Example (1) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0 ABCB BDCAB
- 13. LCS Example (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 A BCB B DCAB
- 14. LCS Example (3) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 A BCB B DC AB
- 15. LCS Example (4) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 A BCB BDC A B
- 16. LCS Example (5) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 0 1 1 A BCB BDCA B
- 17. LCS Example (6) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 0 0 1 0 1 1 A B CB B DCAB
- 18. LCS Example (7) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 A B CB B DCA B
- 19. LCS Example (8) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 1 1 1 2 A B CB BDCA B
- 20. LCS Example (10) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 2 1 1 1 1 1 1 AB C B BD CAB
- 21. LCS Example (11) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 1 2 AB C B BD C AB
- 22. LCS Example (12) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 AB C B BDC AB
- 23. LCS Example (13) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 ABC B B DCAB
- 24. LCS Example (14) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 ABC B B DCA B
- 25. LCS Example (15) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C B Yj B B A C D 0 0 0 0 0 0 0 0 0 0 if ( X i == Y j ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] ) 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 ABC B BDCA B
- 29. Finding LCS j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B
- 30. Finding LCS (2) j 0 1 2 3 4 5 0 1 2 3 4 i Xi A B C Yj B B A C D 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 2 1 1 1 1 2 1 2 2 1 1 2 2 3 B B C B LCS (reversed order): LCS (straight order): B C B (this string turned out to be a palindrome)
- 32. Knapsack problem There are two versions of the problem: (1) “0-1 knapsack problem” and (2) “Fractional knapsack problem” (1) Items are indivisible; you either take an item or not. Solved with dynamic programming (2) Items are divisible: you can take any fraction of an item. Solved with a greedy algorithm .